(1) A composite number a is a positive integer number that is greater than 1 and can be expressed as the product of two smaller positive integer numbers, say b and c. This definition restricts b and c to be greater than 1 and neither b or c can be equal to a but they can be equal to each other, if say a=4.
(2) A prime number p is a positive integer number that is greater than 1 and not a composite number, i.e. it cannot be expressed as the product of two smaller positive integer numbers.
(3) Now, 1 is neither a composite number or a prime number according to the definitions above (if I defined them correctly).
(4) A perfect number P is a positive integer number that equals to the sum of its positive integer divisors, excluding itself.
(5) (Definition of a perfect number extends point (3)) That is, 1 is not a composite number, a prime number, nor a perfect number.
Conclusion: P cannot be 1 nor p. All P are a, but not the other way around.
Is this conclusion correct? What worries me is number 1. We know that division by zero is undefined, therefore we cannot add 1 and 0 and say that 1 is a perfect number. Because if we add zero to 1, then it means that zero is a factor of 1 and we can divide by zero, but we know that we cannot divide by zero. So, 1 cannot be a perfect number.
Best Answer
As commented already by Don Thousand, the definitions are correct.
I think you are confusing the definition of perfect number with $k$-perfect (or multiperfect) number.
So let $\sigma(x)$ be the usual sum of (all) positive divisors of $x$, including $x$.
Perfect numbers, in the traditional sense of the word, are numbers $N$ that satisfy $\sigma(N) = 2N$ (since they ought to satisfy $\sigma(N) - N = N$, per your Definition (4)).
Multiperfect (or $k$-perfect) numbers $M$, on the other hand satisfy $\sigma(M) = kM$.
So the usual perfect numbers are just the $2$-perfect numbers.
Since $\sigma(1)=1$ (because $\sigma$ is multiplicative), then $1$ is $1$-perfect.